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A pea plant is heterozygous for the genes for seed form (Rr) and seed color (Gg). It is crossed with a pea plant of the genotype rrgg. How many of the 60 offspring of this mating are predicted to show the dominant trait for both genes? (The genes are not linked.)Multiple Choice5033150            30

Question

A pea plant is heterozygous for the genes for seed form (Rr) and seed color (Gg). It is crossed with a pea plant of the genotype rrgg. How many of the 60 offspring of this mating are predicted to show the dominant trait for both genes? (The genes are not linked.)Multiple Choice5033150            30

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Solution

To solve this problem, we need to understand the basics of Mendelian genetics.

The first plant is heterozygous for both traits, meaning it has one dominant and one recessive allele for each trait (Rr and Gg). The second plant is homozygous recessive for both traits (rr and gg).

We can use a Punnett square to predict the genotypes of the offspring.

For the seed form (Rr x rr), the possible genotypes are Rr and rr.

For the seed color (Gg x gg), the possible genotypes are Gg and gg.

The offspring that show the dominant trait for both genes must have at least one dominant allele for each trait.

So, the only genotypes that show the dominant trait for both genes are RrGg and Rrgg.

If we multiply the probabilities of these genotypes, we get the total probability of an offspring showing the dominant trait for both genes.

The probability of Rr is 1/2, and the probability of Gg is 1/2.

So, the total probability is (1/2) * (1/2) = 1/4.

If there are 60 offspring, we would expect 1/4 of them, or 15, to show the dominant trait for both genes.

So, the answer is 15.

This problem has been solved

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